Graphs have path connected subsets, namely those subsets for which
every pair of points has a path of edges joining them. But it is not
always possible to find a topology on the set of points which induces
the same connected sets. The 5-cycle graph (and any n-cycle with n>3
odd) is one such example.

I cannot understand this at all. Surely we are talking about putting a topology on the vertex set of the graph. So if we have a path connected graph, the claim seems to be that it is impossible (for some graphs) to find a topology which is also connected. But what's wrong with the indiscrete topology? I would argue that this is a silly topology-- I'd want to topologically distinguish points. But then, if the vertices of the graph are $\{a,b,c,d,\cdots\}$ then try the topology $\tau=\{ \{a,b,c,d,\cdots\}, \{b,c,d,\cdots\}, \{c,d,\cdots\}, \cdots \}$ seems to distinguish points, and be connected.

So maybe the quote implicitly assumes Hausdorff or something? Then how can it possibly only be true for $n$-cycles with "$n>3$ odd"??

1 Answer
1

Given an undirected graph $G = (V, E)$, we would like to know whether there exists a topology $\mathscr{T}$ on the vertices, such that for every $S \subseteq V$, the induced subgraph on $S$ is connected if and only if $S$ is connected w.r.t. $\mathscr{T}$ (i.e., the subspace topology on $S$ is connected). And the claim is that no such topology exists for $G = C_{5}$.

The mistake in the OP is that while you are checking the connectedness of the whole set $V$, you are not checking whether the desired condition is met for not every subset $S \subseteq V$.